**Question**: Determine the length of the segment \( DF \). **Given Information**: - \( BF = 9 \, \text{cm} \) - \( EF = 4 \, \text{cm} \) - \( DF = EF \) **Options**: * \( 5.5 \, \text{cm} \) * \( 5 \, \text{cm} \) (correct answer is circled) * \( 6.5 \, \text{cm} \) * \( 6 \, \text{cm} \) **Diagram**: The diagram displays a circle with center \( F \) and a diameter \( BC \). The points \( D \) and \( E \) lie on the circumference of the circle such that \( DF = EF = 4 \, \text{cm} \). Segment \( DF \) is perpendicular to the diameter and forms two right-angled triangles within the circle. - \( BF = 9 \, \text{cm} \) is indicated as a radius that meets \( DF \) at point \( F \), leading to \( DF = EF \). - The distance from point \( B \) to \( D \) has been divided into the segments, highlighting the triangle relationships used to determine the segment \( DF \). **Explanation**: To find the length of the segment \( DF \), we can use the Pythagorean theorem in triangle \( BDF \): \[ BF^2 = DF^2 + EF^2 \] Given: \[ BF = 9 \, \text{cm}, \quad EF = 4 \, \text{cm} \] The calculation involves finding: \[ DF^2 = BF^2 - EF^2 \] Thus: \[ (9 \, \text{cm})^2 = DF^2 + (4 \, \text{cm})^2 \] \[ 81 \, \text{cm}^2 = DF^2 + 16 \, \text{cm}^2 \] \[ DF^2 = 81 \, \text{cm}^2 - 16 \, \text{cm}^2 = 65 \, \text{cm}^2 \] \[ DF = \sqrt{65} \] Approximating, \(
**Question**: Determine the length of the segment \( DF \). **Given Information**: - \( BF = 9 \, \text{cm} \) - \( EF = 4 \, \text{cm} \) - \( DF = EF \) **Options**: * \( 5.5 \, \text{cm} \) * \( 5 \, \text{cm} \) (correct answer is circled) * \( 6.5 \, \text{cm} \) * \( 6 \, \text{cm} \) **Diagram**: The diagram displays a circle with center \( F \) and a diameter \( BC \). The points \( D \) and \( E \) lie on the circumference of the circle such that \( DF = EF = 4 \, \text{cm} \). Segment \( DF \) is perpendicular to the diameter and forms two right-angled triangles within the circle. - \( BF = 9 \, \text{cm} \) is indicated as a radius that meets \( DF \) at point \( F \), leading to \( DF = EF \). - The distance from point \( B \) to \( D \) has been divided into the segments, highlighting the triangle relationships used to determine the segment \( DF \). **Explanation**: To find the length of the segment \( DF \), we can use the Pythagorean theorem in triangle \( BDF \): \[ BF^2 = DF^2 + EF^2 \] Given: \[ BF = 9 \, \text{cm}, \quad EF = 4 \, \text{cm} \] The calculation involves finding: \[ DF^2 = BF^2 - EF^2 \] Thus: \[ (9 \, \text{cm})^2 = DF^2 + (4 \, \text{cm})^2 \] \[ 81 \, \text{cm}^2 = DF^2 + 16 \, \text{cm}^2 \] \[ DF^2 = 81 \, \text{cm}^2 - 16 \, \text{cm}^2 = 65 \, \text{cm}^2 \] \[ DF = \sqrt{65} \] Approximating, \(
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
![**Question**: Determine the length of the segment \( DF \).
**Given Information**:
- \( BF = 9 \, \text{cm} \)
- \( EF = 4 \, \text{cm} \)
- \( DF = EF \)
**Options**:
* \( 5.5 \, \text{cm} \)
* \( 5 \, \text{cm} \) (correct answer is circled)
* \( 6.5 \, \text{cm} \)
* \( 6 \, \text{cm} \)
**Diagram**:
The diagram displays a circle with center \( F \) and a diameter \( BC \). The points \( D \) and \( E \) lie on the circumference of the circle such that \( DF = EF = 4 \, \text{cm} \). Segment \( DF \) is perpendicular to the diameter and forms two right-angled triangles within the circle.
- \( BF = 9 \, \text{cm} \) is indicated as a radius that meets \( DF \) at point \( F \), leading to \( DF = EF \).
- The distance from point \( B \) to \( D \) has been divided into the segments, highlighting the triangle relationships used to determine the segment \( DF \).
**Explanation**:
To find the length of the segment \( DF \), we can use the Pythagorean theorem in triangle \( BDF \):
\[
BF^2 = DF^2 + EF^2
\]
Given:
\[
BF = 9 \, \text{cm}, \quad EF = 4 \, \text{cm}
\]
The calculation involves finding:
\[
DF^2 = BF^2 - EF^2
\]
Thus:
\[
(9 \, \text{cm})^2 = DF^2 + (4 \, \text{cm})^2
\]
\[
81 \, \text{cm}^2 = DF^2 + 16 \, \text{cm}^2
\]
\[
DF^2 = 81 \, \text{cm}^2 - 16 \, \text{cm}^2 = 65 \, \text{cm}^2
\]
\[
DF = \sqrt{65}
\]
Approximating, \(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ec37281-333e-4c9c-b617-0971eda72364%2Fc7a57530-753a-406a-abfb-2c10f1a6a36a%2Fbiw8zan.jpeg&w=3840&q=75)
Transcribed Image Text:**Question**: Determine the length of the segment \( DF \).
**Given Information**:
- \( BF = 9 \, \text{cm} \)
- \( EF = 4 \, \text{cm} \)
- \( DF = EF \)
**Options**:
* \( 5.5 \, \text{cm} \)
* \( 5 \, \text{cm} \) (correct answer is circled)
* \( 6.5 \, \text{cm} \)
* \( 6 \, \text{cm} \)
**Diagram**:
The diagram displays a circle with center \( F \) and a diameter \( BC \). The points \( D \) and \( E \) lie on the circumference of the circle such that \( DF = EF = 4 \, \text{cm} \). Segment \( DF \) is perpendicular to the diameter and forms two right-angled triangles within the circle.
- \( BF = 9 \, \text{cm} \) is indicated as a radius that meets \( DF \) at point \( F \), leading to \( DF = EF \).
- The distance from point \( B \) to \( D \) has been divided into the segments, highlighting the triangle relationships used to determine the segment \( DF \).
**Explanation**:
To find the length of the segment \( DF \), we can use the Pythagorean theorem in triangle \( BDF \):
\[
BF^2 = DF^2 + EF^2
\]
Given:
\[
BF = 9 \, \text{cm}, \quad EF = 4 \, \text{cm}
\]
The calculation involves finding:
\[
DF^2 = BF^2 - EF^2
\]
Thus:
\[
(9 \, \text{cm})^2 = DF^2 + (4 \, \text{cm})^2
\]
\[
81 \, \text{cm}^2 = DF^2 + 16 \, \text{cm}^2
\]
\[
DF^2 = 81 \, \text{cm}^2 - 16 \, \text{cm}^2 = 65 \, \text{cm}^2
\]
\[
DF = \sqrt{65}
\]
Approximating, \(
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